Question

The following are the observed ABO blood group phenotypes among a group of 600 Canadian Indians.

0=288 A=280 B=19 AB=13

Apply a procedure and calculate the gene frequencies of the alleles involved. Show your work.

Frequencies -

I^a =

I^b =

i =

Answer 1

Given:

Number of individuals with blood group O = 288
Freq(O Blood group) = r2 = 288/600 = 0.48
Number of individuals with blood group A = 280
Freq(A Bood group) = p2 + 2pr = 280/600 = 0.466
Number of individuals with blood group B = 19
Freq(B Blood group) = q2 + 2qr = 19/600 = 0.032
Number of individuals with blood group AB = 13
Freq(AB Blood group) = 2pq = 13/600 = 0.022

Therefore,

r = 0.693
=> p2 + 2pr = 0.466
=> p2 + 2p * 0.693 -0.466 = 0
=> (p - 0.279753) (p + 1.66575) = 0
Thus, p 0.28

So,

p = Freq(IA) = 0.28
q = Freq(IB) = 1 - 0.28 - 0.693 = 0.027
r = Freq(i) = 0.693